Question

Birth and death rates of animal populations typically are not constant; instead, they vary periodically with the passage of seasons. Find P(t) if the population P satisfies the initial value problemdP/dt = (k+bcos2πt)Pwhere t is in years and k and b are positive constants. Thus the growth-rate function r(t)=k+bcos2πt varies periodically about its mean value k. Construct a graph that contrasts the growth of this population with one that has the same initial value P0 but satisfies the natural growth equation P0=kP (same constant k). How would the two populations compare after the passage of many years?

Answer 1

Answer:

We readily separate the variables and integrate:  

∫dP/P=∫(k+bcos2$\pi$t)dt

ln P=kt+(b/2$\pi$)*sin2$\pi$t+ln C

Clearly C = Po, so we find that P(t) = Poexp(kt + (b/2$\pi$)* sin 2$\pi$t). The 271- curve with the typical numerical values P_o = 100, k = 0.03, and b = 0.06. It oscillates about the curve which represents natural growth with P_o and k = 0.03. We see that the two agree at the end of each full year.  

note:

find the attached graph